The shortest distance between lines `barr=(lambda-1)hati+(lambda+1)hatj-(1+lambda)hatk` and
`barr=(1-mu)hati+(2mu-1)hatj+(mu+2)hatk` is

Options

A. `(sqrt(5))/(2)`
B. `(5)/(sqrt(2))`
C. 15
D. `5sqrt(2)`

Answer

To find the shortest distance between the two given lines, we will follow these steps:

### Step 1: Write the equations of the lines in vector form
The first line is given as:
\[ \mathbf{r_1} = (\lambda - 1) \hat{i} + (\lambda + 1) \hat{j} - (1 + \lambda) \hat{k} \]
We can express it in the form:
\[ \mathbf{r_1} = \mathbf{a_1} + \lambda \mathbf{b_1} \]
where:
\[ \mathbf{a_1} = -\hat{i} + \hat{j} - \hat{k} \quad \text{and} \quad \mathbf{b_1} = \hat{i} + \hat{j} - \hat{k} \]

The second line is given as:
\[ \mathbf{r_2} = (1 - \mu) \hat{i} + (2\mu - 1) \hat{j} + (\mu + 2) \hat{k} \]
We can express it in the form:
\[ \mathbf{r_2} = \mathbf{a_2} + \mu \mathbf{b_2} \]
where:
\[ \mathbf{a_2} = \hat{i} - \hat{j} + 2 \hat{k} \quad \text{and} \quad \mathbf{b_2} = -\hat{i} - \hat{j} + \hat{k} \]

### Step 2: Calculate the vector \( \mathbf{a_2} - \mathbf{a_1} \)
Now, we find:
\[ \mathbf{a_2} - \mathbf{a_1} = (\hat{i} - \hat{j} + 2\hat{k}) - (-\hat{i} + \hat{j} - \hat{k}) \]
This simplifies to:
\[ \mathbf{a_2} - \mathbf{a_1} = \hat{i} + \hat{i} - \hat{j} - \hat{j} + 2\hat{k} + \hat{k} = 2\hat{i} - 2\hat{j} + 3\hat{k} \]

### Step 3: Calculate the cross product \( \mathbf{b_1} \times \mathbf{b_2} \)
Next, we compute the cross product:
\[ \mathbf{b_1} \times \mathbf{b_2} = (\hat{i} + \hat{j} - \hat{k}) \times (-\hat{i} - \hat{j} + \hat{k}) \]
Using the determinant method:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ -1 & -1 & 1 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 1 & -1 \\ -1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -1 \\ -1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ -1 & -1 \end{vmatrix} \]
\[ = \hat{i} (1 - 1) - \hat{j} (1 - 1) + \hat{k} (1 + 1) = 0\hat{i} - 0\hat{j} + 2\hat{k} = 2\hat{k} \]

### Step 4: Calculate the magnitude of the cross product
The magnitude of \( \mathbf{b_1} \times \mathbf{b_2} \):
\[ |\mathbf{b_1} \times \mathbf{b_2}| = |2\hat{k}| = 2 \]

### Step 5: Calculate the shortest distance
The shortest distance \( d \) between the two lines is given by the formula:
\[ d = \frac{|(\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2})|}{|\mathbf{b_1} \times \mathbf{b_2}|} \]
Calculating the dot product:
\[ (\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2}) = (2\hat{i} - 2\hat{j} + 3\hat{k}) \cdot (2\hat{k}) = 0 + 0 + 6 = 6 \]
Thus, the shortest distance is:
\[ d = \frac{|6|}{2} = 3 \]

### Final Answer:
The shortest distance between the two lines is \( 3 \).